Nothing cool about the last digits

9 9 9 \LARGE 9^{9^{9}}

What are the first three digits of the number above?

This problem is inspired by bobbym


The answer is 428.

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3 solutions

Let A = 9 9 9 log 10 A = 9 9 log 10 9 A = 9^{9^9} \\ \implies \log_{10} A = 9^9 \log_{10} 9

By computation,

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In[12]:= N[9^9 Log10[9], 32]

Out[12]= 3.6969309963157035874354309509548*10^8

We have,

log 10 A = 9 9 log 10 9 = 3.6969309963157035874354309509548 × 1 0 8 \log_{10} A = 9^9 \log_{10} 9 = 3.6969309963157035874354309509548 \times 10^8

Raising both sides to the power of 10,

A = 1 0 3.6969309963157035874354309509548 × 1 0 8 \implies A = 10 ^ { 3.6969309963157035874354309509548 \times 10^8 }

By computation,

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In[13]:= N[10^(3.6969309963157035874354309509548 10^8), 32]

Out[13]= 4.2812477317574704803698*10^369693099

So, A = 4.2812477317574704803698 × 1 0 369693099 A = 4.2812477317574704803698 \times 10^{369693099}

Moderator note:

What is the difference between your approach, and calculating 9 9 9 9 ^ { 9 ^ 9 } directly? Is one better than the other? If so, why?

It is not apparent to me that finding the accuracy of 9 9 × log 10 9 9^9 \times \log_{10} 9 is easily done. Note that the "initial few digits" is actually determined by the "fractional part of "9^9 log 9".

I differ with your answer Agnishom Chattopadhyay
The value of 9 9 9 9^{9^9} is 9 9 9 = 9 387420489 = 1.9662705 × 1 0 77 9^{9^9} = 9^{387420489} = 1.9662705 × 10^{77}

So the answer for your question i.e., first 3 digits will be 196 \boxed{196} So once have a look.

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You have calculated ( 9 9 ) 9 \left( 9^9 \right) ^ 9 instead.

If you simply threw 9^9^9 into a calculator, that is what it would have given you, because it works from left to right. Remember that tower of exponents is done top down (hence right to left).

Calvin Lin Staff - 6 years ago

How do you calculate that?

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Windows Calculator :P

Digvijay Janartha - 6 years ago

I agree with Chandra Sekhar Vadigalla

Anwesh Pandey - 6 years ago
Hun-Min Park
Jul 24, 2015

(Similar idea to Agnishom Chattopadhyay)

Let N = 9 9 9 = a 1 a n b 1 b m N=9^{9^9}=\overline{a_1 \cdots a_n b_1 \cdots b_m} . Observe that a 1 a n 1 0 m N < a 1 a n 1 0 m + 1 \overline{a_1 \cdots a_n }\cdot 10^m\leq N < \overline{a_1 \cdots a_n}\cdot 10^{m+1} Then we get log 10 a 1 a n + m log 10 N < log 10 a 1 a n + m + 1 \log_{10} \overline{a_1 \cdots a_n }+m\leq \log_{10} N<\log_{10} \overline{a_1 \cdots a_n }+m+1 which implies ( fractional part of log 10 N ) = log 10 a 1 a n ( n 1 ) (\text{fractional part of }\log_{10} N)=\log_{10} \overline{a_1 \cdots a_n } -(n-1)
This makes the calculation faster.

  • Code example (in python):
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from math import log10 as log

def first_n_digits(log_N, n): # log_N = value of log(N)
    return int(10**(log_N-int(log_N)+n-1))

print first_n_digits((9**9)*log(9), 3)

  • Output :
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428

Lu Chee Ket
Dec 9, 2015

Press [ 9 ][ * ][ = ][ = ][ = ]... or [ 9 ][ * ][ * ][ = ][ = ][ = ].. depends onto type of calculator we are using, observe changes of first 3 digits.

There is valuable theory being identified within this week. To tell a hint, first few digits belong to propagating figures while last few digits belong to no squeezed out requirement. The more the better and the least the better respectively but the demand may go against them.

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428124773175749798 [18]
42812477029460678 [17]
427836765123 [12]
218540170 [9]
54359 [5]

We can realized from analysis that 428124... could be the first 6 digits. A range from 9 to 18 is an option of length required for using extract-able 18 S.F. real numbers.

9 9 9 = 1 0 387420489 log 9 9^{9^9} = 10^{387420489 \log 9} = 1 0 369693099.63157035874354309509548 10^{369693099.63157035874354309509548} = 1 0 369693099 × 1 0 0.63157035874354309509548 10^{369693099} \times 10^{0.63157035874354309509548}

= 4.2812477317574704803698711592765 × 1 0 369693099 4.2812477317574704803698711592765 \times 10^{369693099}

Scientific method which is useful is proven by mathematical method.

Answer: 428 \boxed{428}

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